Remark on centrifugal force in Heino Falcke's black hole book

Dear all,

currently I'm reading the Dutch translation of Heino Falcke's "Light in the Darkness: black holes, the universe and us" as a preparation on a course on black hole I'm giving later this year. In part 1 it contains a remark about space telescopes, and the author imagines us to travel with the orbiting telescope. It says in the English translation (from Google books, page number isn't shown there, in the Dutch translation it's page 17)

"In orbit, gravity still has us in its grip but we feel weightless because the centrifugal force and the force of gravity are perfectly balanced."

Am I wrong thinking this passage is confusing? There is only a "balance" between gravity and the centrifugal force because you have to include that fictitious force if you decide riding along with the telescope. That's always the case if you decide to travel with an accelerating object, nothing special about it. It also can't be the common mixing up of centrifugal and centripetal force, because the centripetal force of course isn't "balansed" by gravity as some sort of force equilibrium. The true reason for the orbit is simply because the telescope got the right orbiting velocity and gravity is providing the centripetal force. (I guess the remark about weightlesness is because the orbital speed is too slow to induce any artifical weight, as someone on a merry go around or rotating spaceship would experience). What do you think?

And yes, I know the general attitude about popular science by some members here; I'm just asking whether this remark makes sense.

I agree it's nonsense. In a Newtonian model you are accelerating in a circle, so clearly the net force on you is non-zero. The reason you feel weightless is that the same acceleration (ignoring tidal forces) is applied to every molecule of your body, whereas when you stand on a planet all your squishy parts are hanging from the bony bits.

Ibix said:

I agree it's nonsense. In a Newtonian model you are accelerating in a circle, so clearly the net force on you is non-zero. The reason you feel weightless is that the same acceleration (ignoring tidal forces) is applied to every molecule of your body, whereas when you stand on a planet all your squishy parts are hanging from the bony bits.

Ok, glad I'm not stupid :P I'm opening this topic, because recently I published a book in Dutch about common physics misconceptions, also covering the notorious centrifugal force, and to me this seems one of those common misconceptions ;)

haushofer said:

"In orbit, gravity still has us in its grip but we feel weightless because the centrifugal force and the force of gravity are perfectly balanced."

Am I wrong thinking this passage is confusing? There is only a "balance" between gravity and the centrifugal force because you have to include that fictitious force if you decide riding along with the telescope. That's always the case if you decide to travel with an accelerating object, nothing special about it.

It makes perfect sense to me. If one is asking questios about one's perceptions then adopting the person's accelerating rest frame is appropriate.

Naturally, if this rest frame is accelerating, one must consider the inertial force(s) associated with that acceleration when doing a force balance.

So we do the force balance and see that gravity and centrifugal force are in balance. We are subject to no other external forces. Since both of these are inertial forces, acting on all parts of our body proportionally to mass, our bodies are under no resulting stress. We are in free fall and feel weightless.

This is not the case in an arbitrary accelerating rest frame. If, for instance, we are in a rocket in orbit with the thrusters firing then we have centrifugal force, gravity and the force from the seat on our pants arising from the thrusters. Only two of those three are inertial forces. One of those results in stress on our bodies and the lack of a feeling of free fall.

The reasoning is that if gravity and centrifugal force are perfectly balanced then no other external force exists.

jbriggs444 said:

It makes perfect sense to me. If one is asking questios about one's perceptions then adopting the person's accelerating rest frame is appropriate.

Naturally, if this rest frame is accelerating, one must consider the inertial force(s) associated with that acceleration when doing a force balance.

So we do the force balance and see that gravity and centrifugal force are in balance. We are subject to no other external forces. Since both of these are inertial forces, acting on all parts of our body proportionally to mass, our bodies are under no resulting stress. We are in free fall and feel weightless.

This is not the case in an arbitrary accelerating rest frame. If, for instance, we are in a rocket in orbit with the thrusters firing then we have centrifugal force, gravity and the force from the seat on our pants arising from the thrusters. Only two of those three are inertial forces. One of those results in stress on our bodies and the lack of a feeling of free fall.

The reasoning is that if gravity and centrifugal force are perfectly balanced then no other external force exists.

But if I'm freely falling to earth in a straight line, no air friction but gravity as net force, I'm also weightless. But there is no centrifugal force. And if I'm orbiting fast enough, I'll feel artificial weight, so I'm not weightless anymore (the equivalence principle only holds locally). Strictly speaking you're not weightless on any orbit.

How can centrifugal force play a rôle in me feeling weightless then?

Edit: I think I see your point. Centrifugal force is proportional to mass (if you rotate faster you will leave the orbit) and the telescope doesn't exercise any thrust.

A comparison: If I rotate along with a charge in a magnetic field (an inertial observer states that the centripetal force is provided by the Lorentz force), can I say I'm weightless because the centrifugal force balances the Lorentz force?

jbriggs444 said:

So we do the force balance and see that gravity and centrifugal force are in balance. We are subject to no other external forces. Since both of these are inertial forces,...

In the context of Newtonian physics, gravity is not an inertial force, but the key here is that, just like inertial forces, gravity is a body force that is proportional to mass:
https://en.wikipedia.org/wiki/Body_force

A.T. said:

In the context of Newtonian physics, gravity is not an inertial force, but the key here is that, just like inertial forces, gravity is a body force that is proportional to mass:
https://en.wikipedia.org/wiki/Body_force

So you'd say Falcke is right here?

haushofer said:

But if I'm freely falling to earth in a straight line, no air friction but gravity as net force, I'm also weightless. But there is no centrifugal force.

In the context of Newtonian physics, the free falling frame is an accelerated frame. So in the rest frame of a falling person there is an inertial force -mg (upwards) that balances gravity (just like the inertial centrifugal force does in the rotating common rest frame of an orbiting body and the orbit center).

A.T. said:

In the context of Newtonian physics, the free falling frame is an accelerated frame. So in the rest frame of a falling person there is an inertial force -mg (upwards) that balances gravity (just like the inertial centrifugal force does in the rest frame of an orbiting body).

Yes. I think my confusion comes frome the fact that in Newton's physics you can erase gravity globally by freely falling: in the freely falling frame, there is no gravity. In the orbiting frame this is no longer possible; gravity is not erased globally, but gravity is "counterbalancing" the centrifugal force along the orbit.Hmmm, I'm still a bit confused, I have to think about it a bit.

haushofer said:

So you'd say Falcke is right here?

I don't like his explanation, because it invokes frame dependent effects (inertial forces) to explain frame independent effects (lack of internal stresses, feeling of weightlessness).

The feeling of weightlessness (no internal stresses) can be explained in the inertial frame (without any inertial forces), by the fact that the gravity is a mass-proportional body force, which is locally approx. uniform and thus accelerates all body pieces in the same way.

haushofer said:

in Newton's physics you can erase gravity globally by freely falling: in the freely falling frame, there is no gravity.

Not really, it works only locally, where gravity is approx. uniform.

haushofer said:

In the orbiting frame this is no longer possible; gravity is not erased globally, but gravity is "counterbalancing" the centrifugal force along the orbit.

Don't confuse orbiting frame with rotating frame.
- Inertial centrifugal force only exists in rotating frames. It balances gravity in the rest frame of the orbiting object, that rotates around the center of the orbit.
- In an co-orbiting, but not rotating rest frame of the orbiting object, there is no radial centrifugal force. Just an uniform but time dependent inertial force -ma outwards.

A.T. said:

Not really, it works only locally, where gravity approx. uniform.Don't confuse orbiting frame with rotating frame.
- Inertial centrifugal force only exists in rotating frames. It balances gravity in the rest frame of the orbiting object, that rotates around the center of the orbit.
- In an co-orbiting, but not rotating rest frame of the orbiting object, there is no radial centrifugal force. Just an uniform but time dependent inertial force -ma outwards.

Yes, I assumed a globally uniform gravitational field, but Earth's field of course isn't uniform but radial, you're right. Sloppiness.

I don't get your second remark though. Why is there no radial centrifugal force in a co-orbiting frame? Isn't your "uniform but time dependent inertial force -ma outwards" exactly that?

I'll sit down and do some calculations/transformations to transform away my confusion :P

haushofer said:

TL;DR Summary: A question about how to interpret a remark on centrifugal force in a popular science book.

centrifugal force and the force of gravity are perfectly balanced."

Some rather fanciful language here. This is only true for a circular orbit. A real orbit will be elliptical in which the forces are mostly not balanced and so the radius constantly changes.

This is an entirely separate issue (surely) from the feeling of weightlessness when orbiting in a spacecraft. That sensation is due to the fact that the passenger and the craft are both in free fall so the forces between them are nearly zero..

That is the issue of microgravity; the tiny difference between the forces acting in different parts of the craft. The craft is rigid so parts are going faster than their free fall / orbital speed, if not connected and others are going slower. For objects fixed to the craft, the forces are non zero.

haushofer said:

I don't get your second remark though. Why is there no radial centrifugal force in a co-orbiting frame? Isn't your "uniform but time dependent inertial force -ma outwards" exactly that?

No, because a uniform field is different from a distance dependent radial field (how inertial centrifugal force is usually defined). They might be equal at the COM of the object, but not nearby.

This treatment about the misapplication of the inertial centrifugal force to explain tidal forces goes through various frames of reference, and nicely visualizes the force fields in them:
http://www.vialattea.net/maree/eng/index.htm

It's a popular-science book (one of the better ones, if it's the one I think it is; I've read the German version). However the confusion about inertial forces is notorious in popular-science books and even in the introductory textbook-literature. It becomes a complete mess when it comes to GR and the equivalence principle.

In Newtonian as well as in special relativistic physics you have a class of preferred frames of reference, the inertial frames, where Newton's Lex I holds. All the physical laws are most easily described wrt. to these inertial reference frames, and the Lagrangians describing the dynamics are built from the corresponding symmetry principles (Galilei or Poincare group, respectively).

The description wrt. to other frames of reference are exactly these physical laws. You just transform to the corresponding different coordinates, and there the equations of motion look differently. In Newtonian mechanics it's customary to reinterpret the covariant time derivative on the left-hand side, ##m \ddot{\vec{x}}## as a usual (2nd-order) time derivative of the corresponding components of the position vector and shuffle the rest as "inertial forces" (sometimes called "fictitious forces") to the right-hand side of the equation. If the non-inertial frame is rotating among other pieces you also get the centrifugal force. There's no problem with this approach as long as you stay within this non-inertial frame and don't mix descriptions in inertial and non-inertial frames of reference inadequately.

In GR it's sometimes claimed that "gravity is equivalent to introducing a non-inertial reference frame". This is at best imprecise. The big advance was Einstein's notion that the gravitational interaction cannot be described as a special-relativistic field theory, but that you have to change the spacetime concept quite radically. In modern terms you can understand it as making Poincare invariance local, i.e., there's still the concept of inertial reference frames, but it has only a local meaning.

This leads to the conclusion that the spacetime model for the relativistic description of the gravitational interaction is a pseudo-Riemannian (i.e., Lorentzian) manifold. The local inertial reference frames are given by free-falling non-rotating frames. There the "homogeneous part of the gravitational field" is "gauged away", but you still are not in a global inertial reference frame since you still can observe the "tidal forces". In a sense these are the "true forces" due to the gravitational interaction, and they cannot be gauged away (and they shouldn't, because they are observable effects).

I guess my confusion came from the comparison of a rotating ball attached to a string. If you rotate along with the ball, the tension in the rope equals the centrifugal force. But this tension adapts itself to the rotational speed. So if you rotate faster (without the string breaking), the centrifugal force again balances the tension in the rope. This is no mystery; you decided to rotate along with the ball, so in your rest frame you have to balanse the tension of the string (which you can measure) with the centrifugal force in order to remain still. From a math point of view, this centrifugal force pops up if you transform Newton's 2nd law from inertial to rotating point of view.

The point with the spaceship rotating is of course that at a fixed distance gravity doesn't adapt to the rotation, and if you rotate the ship faster you need extra forces. For gravity this extra force (thrust) will introduce a weight for the observer. From an observer on earth the spaceship and the astronaut both fall at the same rate, hence the astronaut will feel weightless; from the point of view of the astronaut, this is explained by balansing the gravitational force with the centrifugal force the astronaut experiences. But this "balanse" between centrifugal force and gravity is a consequence of the fact that both spaceship and astronaut are falling at the same rate. I think I was confused by reading it more like "the equivalence principe (which is nowhere mentioned; my fault) is a consequence of the fact that the centrifugal force equals gravity in the rotating frame".

It's really fascinating that this can still be so confusing to me. Call it a trauma.

haushofer said:

you decided to rotate along with the ball, so in your rest frame you have to balanse the tension of the string (which you can measure) with the centrifugal force in order to remain still. From a math point of view, this centrifugal force pops up if you transform Newton's 2nd law from inertial to rotating point of view.

Yes, but note that there are different rest frames of the ball you can adapt:
a) rotating around the center of the circle
b) rotating around the ball's COM and translating in a circle around the center with the ball's COM
c) not rotating, just, translating in a circle around the center with the ball's COM (here only the ball's COM is at rest)

The inertial forces in those frame will be different:
a) centrifugal
b) centrifugal + uniform
c) uniform

haushofer said:

I guess my confusion came from the comparison of a rotating ball attached to a string. If you rotate along with the ball, the tension in the rope equals the centrifugal force. But this tension adapts itself to the rotational speed. So if you rotate faster (without the string breaking), the centrifugal force again balances the tension in the rope. This is no mystery; you decided to rotate along with the ball, so in your rest frame you have to balanse the tension of the string (which you can measure) with the centrifugal force in order to remain still. From a math point of view, this centrifugal force pops up if you transform Newton's 2nd law from inertial to rotating point of view.

Of course, that's all correct, but only in the co-rotating non-inertial frame of reference. In the inertial reference frame you have the centriPETAL force on the ball due to the string tension.

haushofer said:

The point with the spaceship rotating is of course that at a fixed distance gravity doesn't adapt to the rotation, and if you rotate the ship faster you need extra forces. For gravity this extra force (thrust) will introduce a weight for the observer. From an observer on earth the spaceship and the astronaut both fall at the same rate, hence the astronaut will feel weightless; from the point of view of the astronaut, this is explained by balansing the gravitational force with the centrifugal force the astronaut experiences. But this "balanse" between centrifugal force and gravity is a consequence of the fact that both spaceship and astronaut are falling at the same rate. I think I was confused by reading it more like "the equivalence principe (which is nowhere mentioned; my fault) is a consequence of the fact that the centrifugal force equals gravity in the rotating frame".

It's really fascinating that this can still be so confusing to me. Call it a trauma.

It's a trauma imposed on all of us by sloppy textbooks ;-)). The best cure of the trauma is a crisp formulation in terms of math!

A.T. said:

Yes, but note that there are different rest frames of the ball you can adapt:
a) rotating around the center of the circle
b) rotating around the ball's COM and translating in a circle around the center with the ball's COM
c) not rotating, just, translating in a circle around the center with the ball's COM (here only the ball's COM is at rest)

The inertial forces in those frame will be different:
a) centrifugal
b) centrifugal + uniform
c) uniform

I guess I have to see the accompanying pictures on the website to see this :P Thanks, I'll check it out!

vanhees71 said:

It's a trauma imposed on all of us by sloppy textbooks ;-)). The best cure of the trauma is a crisp formulation in terms of math!

That's the silly thing: I worked all of this out in my own math-notes a while ago. Apparently doing the calculations didn't cure me. :P

I kept expecting someone here to point out that all of this only applies if your spaceship is not too near a black hole. With GR there is a last stable orbit beyond that, your spaceship will have to be constantly thrusting so it doesn't fall into the black hole. In reality, there is no such orbit, the bremsstrahlung will result in the spaceship's (and the telescope's) orbit decaying until they are inside the last stable orbit.